And you want to find the closed form, of it ( i.e ).
In the Mathematics for the analysis of Algorithms, by Knuth and Greene it has the following procedure:

We construct a function G(z):

Then we multiply (1) by and sum over all n obtaining:

And now it says that we would like to recover the coefficient of in G(z).a) But what does that mean? The only z powers i see in the last equation is , , . So what coefficient of is he talking about?

And then it says that if the denominators of the fractions in G(z) where linear the recovery problem would be simple and that each term would just be a geometric series?b) How come the last one? What exactly does it mean?

And now the most important question, which is about the derivation of the solution.
Book continues saying that since the denominators of the fractions in G(z) are not linear, we express the solution for G(z) in partial fractions so we obtain:(2)

And now it says that we should note that the only nonlinear denominators are just higher powers of a linear factor.
So these terms can be expanded by the binomial theorem and then can be easily computed to be:

c) So my question is: How can be easily computed from (2)???? I don't get it.(Nerd)

Any help with a), b), c) questions and especially with c)?

Thanks in advance.(Thinking)

Jul 4th 2011, 02:55 AM

girdav

Re: Closed-form solution of a recurrence relation.

We expand in power series: ; ; .

Jul 4th 2011, 04:58 AM

ChessTal

Re: Closed-form solution of a recurrence relation.

Quote:

Originally Posted by girdav

We expand in power series: .

Is the second one a typo?
I mean the correct one should have been:
Right?

But even that way, i don't seem to derive the expected solution of:

Here is why:

So what is exactly the whole deal i'm missing? What i'm getting wrong?(Doh)

Jul 4th 2011, 05:04 AM

girdav

Re: Closed-form solution of a recurrence relation.

Yes, the second one was a typo and I have corrected it.
There is a problem in (2): we have whereas with (2) we get . I guess there should be a "-", but I didn't look where. You missed the between the lines 3 and 4 when you write with power series.

Jul 4th 2011, 06:55 AM

ChessTal

Re: Closed-form solution of a recurrence relation.

Quote:

Originally Posted by girdav

Yes, the second one was a typo and I have corrected it.
There is a problem in (2): we have whereas with (2) we get . I guess there should be a "-", but I didn't look where. You missed the between the lines 3 and 4 when you write with power series.

Yes the calculated G(z) in (2) i gave in the initial post** was wrong, there is a minus sign in the last term instead of a positive one i previously had .
So everything makes sense now, thanks.(Rofl)

**I corrected the initial post now.

Jul 4th 2011, 08:29 AM

FernandoRevilla

Re: Closed-form solution of a recurrence relation.

Quote:

Originally Posted by ChessTal

Suppose you have a recurrence relation like: (1) With

If you want to practice, here you have a similar problem I solved to the students of Industrial Engineering .

Jul 4th 2011, 03:06 PM

Also sprach Zarathustra

Re: Closed-form solution of a recurrence relation.

Quote:

Originally Posted by ChessTal

Suppose you have a recurrence relation like:(1)
With

And you want to find the closed form, of it ( i.e ).
In the Mathematics for the analysis of Algorithms, by Knuth and Greene it has the following procedure:

We construct a function G(z):

Then we multiply (1) by and sum over all n obtaining:

And now it says that we would like to recover the coefficient of in G(z).a) But what does that mean? The only z powers i see in the last equation is , , . So what coefficient of is he talking about?

And then it says that if the denominators of the fractions in G(z) where linear the recovery problem would be simple and that each term would just be a geometric series?b) How come the last one? What exactly does it mean?

And now the most important question, which is about the derivation of the solution.
Book continues saying that since the denominators of the fractions in G(z) are not linear, we express the solution for G(z) in partial fractions so we obtain:(2)

And now it says that we should note that the only nonlinear denominators are just higher powers of a linear factor.
So these terms can be expanded by the binomial theorem and then can be easily computed to be:

c) So my question is: How can be easily computed from (2)???? I don't get it.(Nerd)